Let $A$ be a $3 \times 3$ matrix whose elements are only $0$ and $1$. Let $|\lambda_A^{\max}|$ be the spectral radius, i.e., the maximum absolute value of eigenvalues of $A$.
How can I choose the matrix $A$ such that $\log |\lambda_A^{\max}|$ has the minimal positive value?
Can someone give me a hint? Thank you!
I ran the following script on Matlab. Some strange workarounds were necessary because my version is out of date.
The minimum is attained with $$ A = \pmatrix{0&0&1\\1&0&1\\0&1&0}, \qquad \lambda_{max} \approx 1.3247 $$ I'm exploiting the fact that non-negative matrices always have a non-negative eigenvalue which coincides with the spectral radius. We get the same even if we have
L = max(abs(eig(A))).